Answer
$10500$
Work Step by Step
$a_{n+1}−a_n=14+(n+1−1)8−(14+(n−1)8)=8$
As the difference between consecutive terms is constant, the given sequence is arithmetic.
The formula for the finite sum of the arithmetic sequence is:
$S_{n}=\frac{n(a_{1}+a_{n})}{2}$
$a_{n}=14+(n-1)8$
$a_{1}=14+(1-1)8=14$
.
$a_{50}=14+(50-1)8=406$
$S_{50}=\frac{50(a_{1}+a_{50})}{2}=\frac{50(14+406)}{2}=10500$
